University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.4 - The Derivative as a Rate of Change - Exercises - Page 145: 17


a) When the engine stopped, the rocket was climbing at a speed of $190$ ft/sec. b) The engine burnt for $2$ seconds. c) The rocket reached its highest point at the $8$th second, when the velocity was $0$. d) The parachute popped out after about $10.8$ seconds, when the rocket's speed was $90$ ft/sec. e) The rocket fell for $2.8$ seconds before the parachute opened. f) The rocket's acceleration was greatest at the $2$nd second, right before the engine burnt out. g) The acceleration is constant from the $2$nd to the $10.8$th second. Its value is around $-32$ ft/sec^2.

Work Step by Step

a) We see from the graph that at the first $2$ seconds, the velocity increases dramatically. This must correspond to the time the propellant is still burning, accelerating the rocket upward. Then from the $2$th second onward, the velocity keeps dropping constantly for around $9$ seconds. This corresponds to the time the propellant has burnt out, so it cannot push the rocket's velocity anymore. Therefore, the moment the engine stopped is the point where the velocity reaches its maximum, which is at $t=2$ and the rocket was going at a speed of $190$ ft/sec. b) As we claimed in part a), the duration the engine burned is the duration the velocity increases dramatically at first. So the engine burned for $2$ seconds. c) As mentioned in the exercise, after the engine burnt out, the rocket still coated upward for a while. So even though the rocket reached its maximum speed when the rocket burned out, it did not reach its highest point there. So to look for where the rocket's highest point, we look at the velocity again. After the increase, the velocity keeps decreasing; but from $t=2$ to $t=8$, the velocity is still positive, meaning that the rocket still went upward. After $t=8$, the velocity passes the zero point and becomes negative, corresponding to the fact that the rocket now began to fall. Therefore, the highest point must come at $t=8$ second, when the velocity finally reaches $0$ and begins to be negative after that. d) During the time of falling, an explosive charge popped out a parachute to slow down the fall of the rocket. In the graph, this must corresponding to the small rise of the graph after a long fall. This happens at about $t=10.8$ second, at which the speed of the rocket was already $90$ ft/sec. e) The rocket started to fall after its velocity reaches $0$ at $t=8$ second, and the parachute opened at $t=10.8$ second, so the rocket fell for $2.8$ seconds before the parachute opened. f) The acceleration describes the rate of change of the velocity. In the graph, it represents the slope of the tangent at a point. So as the graph is steeper, the acceleration is greater. We see that the graph is steepest during the burning time of the engine. From $t=0$ to $t=2$, the graph is getting steeper and steeper, meaning the acceleration is rising during that time. It reaches its peak just right before the engine burned out at $t=2$ second, which is when the acceleration reaches its greatest. g) The tangent is also itself, so the slope of the tangent to a line does not change as the line heads forward. So the acceleration is constant during the time the graph portrayed is a line. We can easily find that from $t=2$ to $t=10.8$ seconds, the graph portrayed is a line, so the acceleration is constant during this duration. To estimate this constant acceleration, we take the difference of the two endpoints $(2,190)$ and $(10.8,-90)$: $$\frac{-90-190}{10.8-2}=\frac{-280}{8.8}\approx-32ft/sec^2$$
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